inscribed angles on circle

That's basically the problem. I keep getting $\theta=90-\phi/2$. But I have a feeling its not right. What I did was draw line segments BD and AC. From there you get four triangles. I labeled the intersection of BD and AC as point P. From exterior angles I got my answer.

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user60887 We encourage those who ask questions to accept an answer that is helpful. You can accept ONE answer per question, and you can accept an answer by clicking on the $\large ✓$ to the left of the answer you'd like to accept. (You also get two reputation points for each accepted answer.) You can upvote as many answers as you'd like! – amWhy Apr 28 '13 at 1:03

Your answer is correct. The angle subtended at the centre is $2 \theta$, and by using angle sum property and the fact that tangents are perpendicular to the radii at the point of contact, we can obtain $2\theta + \phi = 180 \Rightarrow \theta = 90 - {\phi \over 2}$

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One way would be to let $E$ be the center of the circle. A standard result in geometry tells you that $AEC=2\theta$. And the two sides $AE$ and $CE$ are of equal lengths, and there are right angles at $A$ and $C$, and the sides $AD$ and $CD$ are also of equal lengths. So the triangle $EAD$ is right triangle congruent to $ECD$. One of the angles in that right triangle is $\theta$, so the other is $90^\circ-\theta$. Therefore what you're looking for is $180^\circ-2\theta$.

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You are correct. What makes you second-guess yourself? I think your method is sound. However, note that you are asked to express $\phi$ as a function of $\theta$, so you want

$$\theta = 90 -\dfrac{\phi}{2} \quad \iff \quad \phi = 180 - 2\theta$$

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Nice to see you, again angel here. Unfortunately, I have not been these days as an active one and honestly this makes me annoying. I hope I do the best for next days Amy. :-) – Babak S. Mar 22 '13 at 20:07